2Al+6HCl⇒3H₂+2AlCl₃
<h3>Further explanation
</h3>
Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:
• 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
• 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
• 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
Reaction
Al+HCl⇒H₂+AlCl₃
aAl+bHCl⇒cH₂+AlCl₃
Al, left=a, right=1⇒a=1
Cl, left=b, right=3⇒b=3
H, left=b, right=2c⇒b=2c⇒3=2c⇒c=3/2
the equation becomes :
Al+3HCl⇒3/2H₂+AlCl₃ x2
2Al+6HCl⇒3H₂+2AlCl₃
Answer:
See Explanation
Explanation:
Given that;
N/No = (1/2)^t/t1/2
Where;
No = amount of radioactive isotope originally present
N = A mount of radioactive isotope present at time t
t = time taken
t1/2 = half life
N/1000=(1/2)^3/6
N/1000=(1/2)^0.5
N = (1/2)^0.5 * 1000
N= 707 unstable nuclei
Since the value of the initial activity of the radioactive material was not given, the activity of the radioactive material after three months is given by;
Decay constant = 0.693/t1/2 = 0.693/6 months = 0.1155 month^-1
Hence;
A=Aoe^-kt
Where;
A = Activity after a time t
Ao = initial activity
k = decay constant
t = time taken
A = Aoe^-3 *0.1155
A=Aoe^-0.3465
Answer:
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Explanation:
The molarity of a solution that contains 35.00 g of CuSO4 dissolved in 250.0 mL of water is 0.88M.
<h3>How to calculate molarity?</h3>
The molarity of a solution can be calculated using the following formula:
Molarity = no of moles/volume
According to this question, a solution consists of 35.00 g of CuSO4 dissolved in 250.0 mL of water.
no.of moles of CuSO4 = 35g ÷ 159.6g/mol
no. of moles of CuSO4 = 0.22 moles
Therefore; molarity of CuSO4 solution is calculated as follows:
M = 0.22 ÷ 0.25
M = 0.88M
Therefore, the molarity of a solution that contains 35.00 g of CuSO4 dissolved in 250.0 mL of water is 0.88M.
Learn more about molarity at: brainly.com/question/12127540
6.4mole•64.06g/1mole=409.98g
